 Hello and welcome to the session I am Deepika here, let's discuss the question which says in the following case determine the direction cosines of the normal to the plane and distance from the origin 2x plus 3y minus z is equal to 5. So let's start the solution, the given equation of the plane is 2x plus 3y minus z is equal to 5 thus the direction ratios the normal to the plane are minus 1 therefore the direction cosines of it are 2 over under root of 2 square plus 3 square plus minus 1 square 3 over under root of 2 square plus 3 square plus minus 1 square minus 1 over under root of 2 square plus 3 square plus minus 1 square or 2 over under root of 4 plus 9 plus 1 3 over under root of 4 plus 9 plus 1 minus 1 over under root of 4 plus 9 plus 1 or 2 over under root 14 3 over under root 14 minus 1 over under root 14. Now on dividing the equation of the plane by under root of 14 we get 2 over root 14x plus 3 over root 14y minus 1 over root 14z is equal to 5 over root 14. Now this is of the firm Lx plus My plus Nz is equal to D where Lmn are the direction cosines of the normal to the plane and D is the distance of the plane from the origin here we see that D is equal to 5 over root 14. So the distance of the plane origin is 5 over root 14 hence the direction cosines of the normal to the plane are 2 over root 14 3 over root 14 minus 1 over root 14 and the distance of the plane from the origin is 5 over root 14. So this is the answer for the upper question this completes our session I hope the solution is clear to you bye and have a nice day.